/sci/ - Science & Math

>>6311443
>Prove that there are infinitely many such pairs.

easy, because they they have the same slope and y-intercept


>inb4 give me a hard one

>>6311452
Mathtrolls, just like I feared.

>professional mathematician
so what exactly do you do?

>>6311460
Solve triple integrals.
Make 300k starting.
Things like that.

>>6311443
>obvious homework thread disguised as pretentious math superiority

eat shit, you little faggot.

does 0.999... = 1?

does 1 + 2 + 3 + ... = -12?

>>6311463
>new to /sci/
>knows all the memes

you just outed yourself again, dipshit.

>>6311461
If it's so easy then solve it. I don't know how to do the OP's problem yet but thinking about it I can tell it's hard.

I doubt you'd be able to even start it you dipshit.

>>6311474
Hey there OP, still

a) samefagging
b) using babby-tier psychological subterfuge

to get your homework done? How's that working for you? I dont eat at McDonalds much btw, but I'd like a McChicken and a small Coke.

OP speaking (none of the above posts are mine).

I hoped there would be some intelligent discussion, so that I could enjoy seeing how you all think.

But clearly, at least for now, this forum is full of morons.

>>6311479
>being this much of a faggot

>>6311443
College level mathtrolls. We used to have three threads at any given time arguing about 0.999..=1

>>6311479
Hey OP. Captain Obvious here. I'm going to have to ask you to cease and desist, as being retardedly apparent is my job.

>>6311463
Oh cool. We don't often get actual mathematicians visiting. All we have is zooologist shitposters. Terrible person.

>>6311479
Well, yeah.

>>6311487
>>6311486
>>6311484

Pro Tip: If you are going to samefag, at least change your writing style somewhat. Its also a good idea to wait a little time, especially on a thread that is so completely overlookable, the chances of four people supporting it inside of two minutes is unequivicably zero. Professional mathematician here, btw.

I'm a girl btw

Come on guys, at least try. I know you're all so angry because you find this problem too difficult, but I'll give you a clue.

The next pair is (2,5). You see? 2 divides 5^2+1 = 26, and 5 divides 2^2+1 = 5.

See if you can another pair, and perhaps you'll see a pattern.

>>6311496
>homework thread
>reported

>>6311443
>I am a professional mathematician

So… do you work at Subway or McDonald's?

>>6311504
Someone posts a serious mathematical problem for once instead of dipshit psedomaths 0.999...=1 threads.

Everyone here so fucking moronic they think it's homework. Because only 0.999... = 1 threads count as serious maths, right?

>>6311522
>serious math thread
>obvious homework thread sold under the transparent and deceptive guise of being a test for /sci/ by a 'professional mathematician'

seriously, what are you, 12?

>>6311542
Don't forget the blatant samefagging.

>>6311542
What the fuck are you? Some high school drop out? My question is so far beyond the limits of your intellect that you don't even recognise it.

>>6311553
Man, reverse psychology thru derogatory insults has already failed you. Get a new strategy. I suggest one of the following:

a: actually paying attention in class and studying
b: googling a math help forum and posting there
c: googling proof by induction

You dont even deserve c, but I am tired of your pathetic retard level attempts at manipulation.

>>6311553
Oh and to answer your other question, I am 36 years old, I have a M.Sc. in Nanotech, and I often answer well-posed homework questions even if they are against the rules, because I spent most of my academic career teaching people.

You shit all over me with your lies tho, so fuck off regarding getting any more help than what I provided above.

>>6311570

a) My OP was phrased politely, if in the spirit of friendly competition.

b) For the millionth time, this is not homework. It is a problem that I spent several hours this evening solving, and was interested to see how other people approached it, and whether anyone had a new solution that was different to mine.

c) I'm 26 and I have a PhD in maths. It's me who should be helping you with your homework.

>>6311583
d) I am full of shit.

>>6311583
>It is a problem that I spent several hours this evening solving

>I have a PhD in maths

Big lels heard by Voyager II

>>6311592
Test me then. Ask me any difficult question that *you* know the answer to (obviously not something stupid like the Goldbach conjecture). I will answer you here and prove that I am no amateur stuck on his homework.

>>6311605

I'm not the guy fucking with you, but how do I use the existence and uniqueness theorem to prove that a first ordinary differential equation
x`(x) = f(x,t) , x(t0)=x0 is bounded

>>6311605
OK then, you asked for it. My question is as follows:
----------------------------------------------------------------------
There are some pairs of integers (n,m) such that both n divides m^2+1 and m divides n^2+1.

The pair (1,2) is an obvious example since 2 divides 1^2+1 = 2, and clearly 1 divides 2^2+1 = 5.

Prove that there are infinitely many such pairs.

>>6311617
Im gonna throw in the extra caveat that there be no remainder in either division, of course.

>>6311614
Do you mean to show that the solution of

<span class="math">\frac{dx}{dt}=f(x,t) \; , \; x(t_0)=x_0[/spoiler]

is bounded?

>>6311619

yes exactly

>>6311619
<googling intensifies>

Can someone please post the answer to OP's problem?

I have a BSc in math and I spent more than an hour failing to solve it.

inb4 samefag

At this point I'm even willing to pay money for a solution. My brain can't take it anymore.

Suppose (n,m) is the largest such pair. Then (nm,m) fulfils the condition, and we have a contradiction. Therefore there is an infinite number of such pairs.

Crossboarding from /pol/. You won't find intelligent discussion on /sci/

>>6311621
It's not. A trivial counterexample would be <span class="math">f(x,t)=1[/spoiler] and <span class="math">x(0)=0[/spoiler]. Then <span class="math">x(t)=t<span class="math"> which is obviuosly not bounded unless you restrict the range of t.[/spoiler][/spoiler]

>>6311637
and /pol/ proves yet again how dumb they really are, maybe you should go back to your eugenics threads and fear the black man?

(2,5) is a pair. (2*5,5) = (10,5) is not a pair.

>>6311443
maybe this will help you with your homework op
http://jeremykun.com/2013/03/21/methods-of-proof-induction/

>>6311637
It's very nice to see at least an intelligently phrased mathematical sentence. But your the implication (n,m) is a pair -> (nm,m) a pair is wrong.

Suppose (n,m) = (2,5). Then 2 div 5^2+1=26 and 5 div 2^2+1=5 so these are clearly a pair.

But you need nm divides m^2+1, which in this case would mean 10 div 26.

Keep trying!

>>6311617
the silence on this one is deafening. well played, anon.

>>6311651
Are you the OP? Do you know the solution?

Please post it. I can't fucking take it anymore. It's hurting my brain.

>>6311651
Answer my fucking question, already.
>>6311617

>>6311660
OP doesnt actually know it. You are posting in a homework thread.

>>6311661
If that's homework, OP's teacher is a major douchebag. I have a maths BSc and I can't solve this. My guess is that it's an unsolved conjecture and OP is trolling.

No idea if troll or not, but I reall don't care, I found the question interesting, and it checks out for every pair of Fibonacci numbers F_n, F_(n+2), like 2 and 5, 5 and 13, 13 and 34, 34 and 89... so I'm assuming that's the answer. Is there any possibility to prove it just working with the very definition of the Fibonacci sequence or are you gonna need some further identities?

>>6311663
Thats also a good guess.

>>6311653
OK, fine. Here's the solution. Define the sequence of integers <span class="math">(x_n)[/spoiler] recursively by <span class="math">x_0=1[/spoiler], <span class="math">x_1=2[/spoiler] and <span class="math">x_{n+1}=\frac{x_n^2+1}{x_{n-1}}[/spoiler].

Then <span class="math">(x_n,x_{n+1})[/spoiler] is always a pair. I won't waste my time proving this, I'll leave it open for someone else to try.

Note that this sequence yields (1,2), (2,5), (5,13) as the first three pairs. You can keep going and they all work.

>>6311665
>I wont waste my time proving this

but that is exactly what you need to do, OP, or you have proven nothing in this thread.

>>6311665
That's what I already found out on my own. Now prove it, you goddamn asshole.

>>6311674
Actually, you're suggesting that the Fibonacci sequence always works. This is different to my sequence, as Fib is defined <span class="math">x_{n+1}=x_n+x_{n-1}[/spoiler]. Let me think a moment to see if this works too.

>>6311674
Look at the recursive definition again. The proof is explicit.

>>6311673
I'm not trying to prove anything. I'm trying to encourage an intelligent math debate. And the last few posters i.e. Fibonacci man have shown that there are some actual thinkers amid all the trolls, which is nice.

>>6311681
I'm not the poster who suggested fibonacci. Can you post the proof now?

>>6311686
>Test me then. Ask me any difficult question that *you* know the answer to (obviously not something stupid like the Goldbach conjecture). I will answer you here and prove that I am no amateur stuck on his homework.

>>6311682
>The proof is explicit.

I can't do it. Please help.

>>6311692
No you are right, sorry. OP's recursive defintion does not guarantee integer results at every step. He does need to prove that.

Alright, OP is a douchebag.

I'll give 5 bitcoins to the first poster who can come up with a complete proof.

>>6311696
Indeed. This is part of the induction proof. I want to show that if <span class="math">(x_{n-1},x_n)[/spoiler] is a pair, then <span class="math">(x_n,x_{n+1})[/spoiler] is a pair.

As you correctly point out, I first need to show that <span class="math">x_{n+1}[/spoiler] is an integer. Fortunately, by induction hypothesis I have <span class="math">x_{n-1}[/spoiler] divides <span class="math">x_n^2+1[/spoiler], which confirms that <span class="math">x_{n+1}[/spoiler] is an integer.

It is much harder to show that <span class="math">(x_n,x_{n+1})[/spoiler] is a pair.

>>6311696
yeah he is basically just restating the original problem more formally. and then he tries to slip out of the proof by calling it a waste of time and, once again, 'leaving it to anon' to show as a test of their intelligence.

>>6311710
what? no. you still havent shown that the sequence will always give integers, it is quite easy to imagine it moving into irrationals for example.

once you prove THAT, the recursive defintion shows easily that every pair in the sequence is a solution.

>>6311666
Regarding Fibonacci: sorry, I misread your post. You are claiming that <span class="math">(F_{2n},F_{2(n+1)})[/spoiler] is always a pair.

Yes, this is correct, and the proof of this would be entirely equivalent to proving that my sequence works, as they are the same!

The equivalence of our sequences follows from something called 'Catalan's identity', a special case of which is that <span class="math">F_{2(n+1)}=\frac{F_{2n}^2+1}{F_{2(n-1)}}[/spoiler]

>>6311712
I am not just restating the problem. First I gave some examples, then I actually stated the solution as a sequence! That's the first big step out of the way. Now what remains is to prove that the sequence works...

>>6311708
And here we have the ultimate proof that OP is full of shit.

If he really knew the answer, he would have posted it by now, considering that 5 BTC is worth $2000.

Thanks for playing, asshole. You blew your cover.

>>6311729
That is all part of the induction proof. P.S. It is impossible to imagine this kind of sequence moving into irrational numbers, as it is defined recursively using only basic arithmetic operations. The worst that would happen is that we'd get fractions. Which we don't, by induction.

>>6311710
>Fortunately, by induction hypothesis

Nope. Your formula uses division to define the next term. You would have to use strictly mutliplication and/or addition to ensure the next term is an integer from the definition alone.

>>6311443
>Translation: Please do my homework for me. Also I am a huge faggot, please rape my face.

>>6311743
>mfw OP can't into integers.

IF A and B are integers, then

C = A + B is an integer and
C = A - B is an integer and
C = A*B is an integer BUT
C = A/B is NOT necessarily an integer.

>>6311744
Not necessarily. Look: <span class="math">\frac{5^2+1}{2}=\frac{25+1}{2}=\frac{26}{2}=13[/spoiler]. Next step, <span class="math">\frac{13^2+1}{5}=\frac{169+1}{5}=\frac{170}{5}=34[/spoiler].

It's a special sequence. It always returns an integer. This follows whenever the previous two elements in the sequence are a pair.

>>6311737
>what is 'more formally'

you are restating the problem, OP. and you dont know the answer.

>>6311754
Sorry, got my formulas wrong. Can't be bothered to type again. Short answer, C=A/B is not necessarily an integer, no, but it doesn't mean it's never an integer. This is a special sequence with special properties. One of them is that every element is an integer.

>>6311751
C=A/B is an integer if B divides A.

OP's proof is correct, modulo obvious pedantic details.

>>6311762
circular fucking argument. jesus christ. and hello, OP.

>I am a professional mathematician

Let X be a compact space and T:X-->F a continuous family of self-adjoint Fredholm operators. What can you say about the index bundle of T?

>>6311758
PROVE IT.

Well OP, me and my 5 bitcoin are waiting for you to post the proof. Do you really not want the money?

>>6311765
>continuous family of self-adjoint Fredholm operators

That there is a bijection between its Grothendieck group act the homotopy classes of the mapping?

>>6311775
You have precisely 0 bitcoin.

>>6311778
Nope, the answer is much simpler.

>>6311781
I posted the screenshot of my wallet. Keep making excuses, dumbfuck. I know you don't have a proof.

>>6311785
>keep on samefagging

>>6311768
Not OP, and not buying that anyone here is so dense as you are pretending to be.

>>6311785
Come on OP. Its not at all apparent. If you didnt add the 1 in, then sure, it would be obvious. But that little 1... it really fucks things up, doesnt it? And you cant prove it.

Is it an unsolved conjecture? Link pls.

OK, I'm going to post as much of the proof as I can without spoiling it completely. Remember that the first pair is <span class="math">(x_0,x_1)=(1,2)[/spoiler] and that we define <span class="math">(\ast) x_{n+1}=\frac{x_n^2+1}{x_{n-1}}[/spoiler] We assume, for the induction hypothesis, that <span class="math">(x_{n-1},x_{n})[/spoiler] is a pair, that is

a) <span class="math">x_{n}[/spoiler] divides <span class="math">x_{n-1}^2+1[/spoiler]

b) <span class="math">x_{n-1}[/spoiler] divides <span class="math">x_{n}^2+1[/spoiler]

We want to show that

c) <span class="math">x_{n+1}[/spoiler] divides <span class="math">x_{n}^2+1[/spoiler]

d) <span class="math">x_{n}[/spoiler] divides <span class="math">x_{n+1}^2+1[/spoiler]

Well, the first part is easy, because by definition <span class="math">x_n^2+1=x_{n+1}x_{n-1}[/spoiler] and <span class="math">x_{n+1}[/spoiler] obviously divides <span class="math">x_{n+1}x_{n-1}[/spoiler].

So it remains to prove part d). Can you do it?

>>6311808
>implying d wasnt the only thing you needed (and were asked) to prove this whole fucking time

STILL trying to push the work off onto anon. STILL dont have the answer. OP confirmed for faggot.

>>6311808
>Can you do it?

I can't and neither can you.

>>6311808
I'll give you money for the fucking prove and I mean that seriously.

>>6311744
Nope. You can verify that the first two elements of the sequence are integers and satisfy the requested relation (are a pair). And because of the definition of the pair, you can show that if <span class="math">(x_{n-1}, x_n)[/spoiler] is a pair then <span class="math">x_{n+1}[/spoiler] is an integer.

Another way to think about it is that "m divides n^2+1" is another way of saying "am = n^2+1" for some integer <span class="math">a[/spoiler]. You can trivially use the existence of that integer <span class="math">a[/spoiler] to prove that the sequence in >>6311665 remains integral.

Nice puzzle OP.

>>6311808
How much money do you want for the proof?

If OP posts the full proof, I will an hero on camera.

>mfw I know he doesn't have it

>>6311465
Not him, but:
0.9999 is just a longer way to write 1, same as 2413.25999999 is another way to write 2413.26
Think of it like:
1 - 0.000000 (and zeroes going on forever, there's no 1 at the end, in fact - there's no end)
= 1 - 0 = 1

Also sum of natural numbers equalls -1/12, but it's a little tricky and you're dealing with infinity. I think it relies on analytic continuation, but I'm not an expert.
As a sidenote - this result (-1/12) is used in physics, namely string theory.

prove there are infinitely many numbers which are divisible by 3 aswell 4

>>6311871
>0.9999 is just a longer way to write 1

No, they're distinct numbers. Writing 1 instead of 0.999... is just laziness. Engineers can do that because the approximation error is small enough. But mathematicians should be rigorous.

I think I have it:

Suppose:
H1) A and B are positive integers.
H2) A<B
H3) A and B are relatively prime
H4) A divides B^2+1
H5) B divides A^2+1

Now let C=(B^2+1)/A. Then

R1) B<C. Proof: C=(B^2+1)/A > B^2/A > AB/A=B
R2) C divides B^2+1. Proof: C=(B^2+1)/A implies A=(B^2+1)/C.
R3) B and C are relatively prime.
Proof: AC=B^2+1. Suppose p is prime and divides both B and C. Then p
divides B^2+1, and thus it does not divide B^2, and thus not B, a
contradiction.

R4) B divides C^2+1
Proof: Let P=(A^2+1)/B. This is an integer by H5. Then C^2+1=
(B^2+1)^2/A^2+1=(B^4+2B^2+1+A^2)/A^2= (B^4+2B^2+PB)/A^2 =
B(B^3+2B+P)/A^2. Since A and B are relatively prime, we must have
A^2|(B^3+2B+P). Let Q=(B^3+2B+P)/A^2. Then C^2+1=BQ and we are done.

Now we can set A=B, B=C, and we have our hypotheses again.

Your a professional mathemagician? Pls tell me how 2 divide by 0. I need to know.

Been a while since my algebra class, but i bet you could tackle this with some modular arithmetic and gcd. You know that m^2=-1 mod n and n^2=-1 mod m. And we can write g=an+bm where g is gcd of n,m and a,b are integers

>>6311914
Not feeling the love here /sci. Y'all demanded a proof, I give one, then nothing. No "attaboy"? No "kudos"? No "nice"? Not even a "that's wrong you retard"?

>>6312079
it is indeed a correct proof, OP is probably gone since this thread was started like 6 hours ago.

I dunno doesn't seem like it was *that* hard of a problem that OP was justified in using it as a litmus test for those versed in mathematics beyond >college level

the subject isn't crazy, it's basic number theory

it's a standard problem type at my undergrad institution, just find a recurrence relation, the naive one in this case, and then prove what needs to be proved about the terms given by the recurrence relation


not trying to take away from your actually proving, good job, i'd rather this thread be about recent advances in sieve methods than OP being a dick and looking down on everyone

>>6312131
Thank you! I appreciate the feedback. I was happy to have gotten it because I am pretty weak with number theory. It wasn't that hard... once I saw that the proof could be completed if gcd(A,B)=1, it was straightforward.

>>6312141
Yea I'm not very strong in number theory either, more of continuous groups and geometry type, not that there aren't overlaps with number theory, but my undergrad advisor once joked about number theorists being numerologists, I think that reflects how willing I am to put time into proving results in the subject.

The point being, good job seeing through the problem, especially since it's in a subject you're not big on.

The equation fails for large numbers.

I would say solve the good will hunting problems if they werent easy. Im not a mathfag so I dont care if you are legit.

Fails before step 2178309. Only by a few digits but it accelerates after that.

The fibonacci sequence and x^2+1 do not share the same set of numbers that make up its modular space at its first digit. This would've been a clue.

>>6312349
No, wait, I'm full of shit.
It has nothing to do with the modular space.
BOY AM I TIRED

Still fails though.

>>6312131
I read through this thread, and while you say this problem is just standard, I'd like to point out that nobody here actually solved it completely.

What seems like the OP actually gave the recurrence relation as a clue, and then half of the induction proof. All /sci/ contributed was an incorrect attempt at completing the induction proof.

So a standard problem maybe, and OP a dick maybe, but this was still too hard for /sci/

http://en.wikipedia.org/wiki/Mean_value_theorem#Cauchy.27s_mean_value_theorem

off topic, but can someone explain cauchy's mean value theorem to me?
where are the two functions f and g defined in the picture on the right?

>>6312555
Lel, you're still there OP?

>>6312558
From a geometric point of view, the equality means that any representative curve of a function f:R->R^2 (f being differentiable) admits a parallel tangent to one of its chords.

>>6312565
thanks for your answer

>f:R->R^2
What does this mean exactly? A real valued function I map onto a plane? How can I make this vivid?

also I don't quite understand why it talks about functions f and g, but in the image there's only one function/graph?

>>6312576
A function <span class="math">\mathbb{R}^2\rightarrow\mathbb{R}^2[/spoiler] is just a curve in the plane. The functions f and g are just the first and second co-ordinates of this curve. So the real number x maps to the co-ordinate <span class="math">(f(x),g(x))[/spoiler].

>>6312586
Sorry, I meant a function <span class="math">\mathbb{R}\rightarrow\mathbb{R}^2[/spoiler]

>>6312576
Ahahah, you are stupid :)

>>6312586
ah, now I get it

thanks a bunch :3

>>6312598
not knowing the definition of anything hardly correlates with intelligence - unlike the fallacy which you just drew

Anyone up for a REAL math challenge?

Here we go:
Find a real life application of OP's theorem!

Without applications number theory is just useless abstract puzzles and no better than philosophy.

>>6312603
It got me into your mum's pants. Is that enough for you?

>>6312555
>but this was still too hard for /sci/
Not true. Complete solution is here
>>6311914

Similar analysis shows that the sequence (1,1), (1,2), (2,5), ... gives
all such pairs. So there is uniqueness.

Suppose B and C are positive integers such that B|(C^2+1) and C|(B^2+1).
Then:
1) B and C are relatively prime. Proof: Suppose n>0 divides B and C.
Then n divides C^2+1 since B does. Then n divides both C^2 and C^2+1, so
n=1.
2) If B=C, then B=C=1. Proof: If B=C, then B=C=1 since gcd(B,C)=1.

Assume now that B<C, and set A=(B^2+1)/C>0. A is an integer by
hypothesis.

3) A<=B, with equality only if A=B=1 and C=2. Proof: A=(B^2+1)/C =
B^2/C+1/C < BC/C+1/C = B+1/C <= B+1/2. Since A and B are integers,
A<=B. If A=B, then BC=AC=B^2+1, so C=B+1/B. Since C is an integer,
A=B=1, and C=CA=B^2+1=2.

4) A and B are relatively prime. If n>0 divides A and B, then n divides
B^2 and CA=B^2+1. So n=1.

5) A|(B^2+1). Proof: CA=B^2+1.
6) B|(A^2+1). Proof: A^2+1=(B^4+2B^2+1+C^2)/C^2=(B^4+2B^2+PB)/C^2
=B(B^3+2B+P)/C^2. Since gcd(B,C)=1, C^2|(B^3+2B+P).

So suppose we are given an initial pair (B,C), with B<=C. If B=C, then
(B,C)=(1,1) and we are done. Otherwise B<C, and we get a new pair (A,B)
with A<B unless (A,B,C)=(1,1,2). If A<B, set C=B and A=B, and continue.
Since A is a positive integer, this process must terminate.

>Can anyone prove themselves to me by solving this problem:

Why dont you just crate a thread about a topic you want to talk about and see if you get some decent answer- oh yeah I forgot that you are too stupid to do your own homework.

>>6312705
This is wrong. The pairs stop working eventually.

>>6312896
provide some proof of that claim

i've read the proof, it's valid, the pair (1,2) works so must all generated from it by the recurrence relation, that's just kind of how induction works.

Whatever you're using to calculate is probably just flawed.

>>6311924
You take the calculus of x.

>>6312896
Where does it stop working? I suspect your computer is dropping digits.

>>6312935
2178309^2+1 is not divisible by 832040
832040^2+1 is not divisible by 2178309

2971215073^2+1 is not divisible by 1134903170
1134903170^2+1 is not divisible by 2971215073

Furthermore, the ratio between each number converges towards ~1+phi.

>>6313132
that doesn't prove anything, how did you get to the number 2178309? you probably got messed up on your way there

>>6313132
>2971215073^2+1 is not divisible by 1134903170
Yes it is.
(2971215073^2+1)/1134903170=7778742049

>1134903170^2+1 is not divisible by 2971215073

Yes it is. (1134903170^2+1)/2971215073=7778742049

(1,2)
(2,5)
(5,13)
(13,34)
(34,89)
(89,233)
(233,610)
(610,1597)
(1597,4181)
(4181,10946)
(10946,28657)
(28657,75025)
(75025,196418)
(196418,514229)
(514229,1346269)
(1346269,3524578)
(3524578,9227465)
(9227465,24157817)
(24157817,63245986)
(63245986,165580141)
(165580141,433494437)
(433494437,1134903170)
(1134903170,2971215073)
(2971215073,7778742049)
(7778742049,20365011074)
(20365011074,53316291173)
(53316291173,139583862445)
(139583862445,365435296162)
(365435296162,956722026041)
(956722026041,2504730781961)

>>6313195
>yfw when anon almost certainly used a simple algorithm to calculate those values
>yfw it is SO easy to check
>yfw I was right the whole time about you not having any proof, and this being a bullshit thread where you try to get other people to do your work by being a pretentious faggot.

>>6313132
I think you are mixed up.
(2971215073^2+1)/1134903170=7778742049

Check against these:

(1,2)
(2,5)
(5,13)
(13,34)
(34,89)
(89,233)
(233,610)
(610,1597)
(1597,4181)
(4181,10946)
(10946,28657)
(28657,75025)
(75025,196418)
(196418,514229)
(514229,1346269)
(1346269,3524578)
(3524578,9227465)
(9227465,24157817)
(24157817,63245986)
(63245986,165580141)
(165580141,433494437)
(433494437,1134903170)
(1134903170,2971215073)
(2971215073,7778742049)
(7778742049,20365011074)
(20365011074,53316291173)
(53316291173,139583862445)
(139583862445,365435296162)
(365435296162,956722026041)
(956722026041,2504730781961)

>>6313367
yes, that's about the size of it. Turned out to be a nice problem though, I think

>>6313367
rage more muh greentext bro, the post you were replying to, mine, is not a post by the OP

>>6311914
I like this numbered hypothesis and result format for writing theorems. I hope you don't mind if I start using it as well.

>>6311765

I consider myself to be on my way to a professional mathematician, and I have no idea what an index bundle is. Math is a big field.